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Q1.

In an isosceles triangle $ABC$ where $AB=AC$, the bisectors of angles $B$ and $C$ meet at $O$. Join $A$ to $O$.
Prove that:
(i) $OB=OC$
(ii) $AO$ bisects $\mathrm{\angle }A$

Solution:

Given: $AB=AC$ and $OB,OC$ are angle bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$.
To prove: (i) $OB=OC$, (ii) $AO$ bisects $\mathrm{\angle }A$.

Proof:

• In $\mathrm{△}ABC$, since $AB=AC$, we have $\mathrm{\angle }B=\mathrm{\angle }C$

• $\mathrm{\angle }OBC=\frac{1}{2}\mathrm{\angle }B$ and $\mathrm{\angle }OCB=\frac{1}{2}\mathrm{\angle }C$.
  ⇒ $\mathrm{\angle }OBC=\mathrm{\angle }OCB$ because $\mathrm{\angle }B=\mathrm{\angle }C$.

• Hence, in $\mathrm{△}OBC$, two angles are equal, so the sides opposite to them are also equal.
  ⇒ $OB=OC$
  

Now, in triangles $OAB$ and $OAC$:

• $OB=OC$ (proved)

• $AB=AC$ (given)

• $AO$ is common.

Therefore, $\mathrm{△}OAB\cong \mathrm{△}OAC$ by SSS.
⇒ $\mathrm{\angle }BAO=\mathrm{\angle }CAO$
Thus, $AO$ bisects $\mathrm{\angle }A$

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Q2.

In $\mathrm{△}ABC$, $AD$ is the perpendicular bisector of $BC$.
Prove that $\mathrm{△}ABC$ is isosceles, i.e. $AB=AC$

Solution:

Given: $AD\perp BC$ and $BD=DC$
To prove: $AB=AC$

Proof:
In $\mathrm{△}ABD$ and $\mathrm{△}ACD$:

• $BD=DC$ (given)

• $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$ (perpendicular)

• $AD$ is common.

So, by RHS congruence (Right angle–Hypotenuse–Side),
$\mathrm{△}ABD\cong \mathrm{△}ACD$
Hence, $AB=AC$

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Q3.

In an isosceles triangle $ABC$, $AB=AC$.
Altitudes $BE$ and $CF$ are drawn from $B$ and $C$ to the opposite equal sides $AC$ and $AB$.
Prove that $BE=CF$.

Solution:

Given: $AB=AC$, $BE\perp AC$, and $CF\perp AB$.
To prove: $BE=CF$

Proof:
In $\mathrm{△}ABE$ and $\mathrm{△}AFC$:

• $AB=AC$ (given)

• $\mathrm{\angle }AEB=\mathrm{\angle }AFC={90}^{\circ }$

• $\mathrm{\angle }BAE=\mathrm{\angle }CAF$ (common angle at A).

By A–A–S congruence,
$\mathrm{△}ABE\cong \mathrm{△}AFC$
Hence, $BE=CF$

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Q4.

In $\mathrm{△}ABC$, $BE$ and $CF$ are equal altitudes from $B$ and $C$ respectively on sides $AC$ and $AB$.
Prove that:
(i) $\mathrm{△}ABE\cong \mathrm{△}ACF$
(ii) $AB=AC$

Solution:

Given: $BE=CF$ $BE\perp AC$ $CF\perp AB$
To prove: (i) $\mathrm{△}ABE\cong \mathrm{△}ACF$ (ii) $AB=AC$

Proof:
In $\mathrm{△}ABE$ and $\mathrm{△}ACF$:

• $BE=CF$ (given)

• $\mathrm{\angle }AEB=\mathrm{\angle }AFC={90}^{\circ }$

• $\mathrm{\angle }BAE=\mathrm{\angle }CAF$ (common).

So, by A–A–S,
$\mathrm{△}ABE\cong \mathrm{△}ACF$
Hence, $AB=AC$
Thus, $\mathrm{△}ABC$ is isosceles.

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Q5.

$\mathrm{△}ABC$ and $\mathrm{△}DBC$ are two isosceles triangles on the same base $BC$
Prove that $\mathrm{\angle }ABD=\mathrm{\angle }ACD$

Solution:

Given: $AB=AC$ and $DB=DC$
To prove: $\mathrm{\angle }ABD=\mathrm{\angle }ACD$

Proof:
Join $AD$
In $\mathrm{△}ABD$ and $\mathrm{△}ACD$:

• $AB=AC$ (given)

• $BD=CD$ (given)

• $AD=AD$ (common).

So, $\mathrm{△}ABD\cong \mathrm{△}ACD$ by SSS congruence.
Hence, $\mathrm{\angle }ABD=\mathrm{\angle }ACD$

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Q6.

$\mathrm{△}ABC$ is isosceles with $AB=AC$. Side $BA$ is produced to $D$ such that $AD=AB$
Prove that $\mathrm{\angle }BCD={90}^{\circ }$

Solution:

Given: $AB=AC$, $AD=AB$
To prove: $\mathrm{\angle }BCD={90}^{\circ }$

Proof:
Since $AB=AC$
$\mathrm{\angle }ABC=\mathrm{\angle }ACB$
Also, since $AD=AB$
$\mathrm{△}ACD$ is isosceles ⇒ $\mathrm{\angle }ADC=\mathrm{\angle }ACD$

Now, in the straight line $AD$,
$\mathrm{\angle }ADC+\mathrm{\angle }ADB={180}^{\circ }$
But $\mathrm{\angle }ADB$ is external to $\mathrm{△}ABC$, so
$\mathrm{\angle }ADB=\mathrm{\angle }ABC+\mathrm{\angle }ACB=2\mathrm{\angle }ACB$
Hence,

$$\mathrm{\angle }ADC={180}^{\circ }-2\mathrm{\angle }ACB\mathrm{}$$

But in $\mathrm{△}ACD$,

$$\mathrm{\angle }ADC+2\mathrm{\angle }ACD={180}^{\circ }\mathrm{.}$$

Substituting $\mathrm{\angle }ADC={180}^{\circ }-2\mathrm{\angle }ACB$, we get

$${180}^{\circ }-2\mathrm{\angle }ACB+2\mathrm{\angle }ACD={180}^{\circ }\mathrm{}$$

⇒ $\mathrm{\angle }ACD=\mathrm{\angle }ACB={45}^{\circ }\mathrm{}$
Therefore, in $\mathrm{△}BCD$

$$\mathrm{\angle }BCD={180}^{\circ }-(\mathrm{\angle }ACB+\mathrm{\angle }ACD)={180}^{\circ }-({45}^{\circ }+{45}^{\circ })={90}^{\circ }\mathrm{}$$

✅ Hence, $\mathrm{\angle }BCD={90}^{\circ }$